Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 11H / b

The continued-fraction algorithm givesafter which the process repeats. ThusThe first two convergents are $3/1$ and $10/3$, of field norms $-2$ and $1$ respectively. The even period gives the fundamental solutionIt is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding Pell equation group law givesIndeed $3970^2-11\cdot 1197^2=1$.